Post Your Comment

62 Comments

Interesting article. Not only because of the look at the new ''low''power Q CPUs. But because of the ideal of using a greater wattage CPU (such as the i920 etc) that utilizes lower total voltage. Although the 'average'consumption is less with a power saving cpu.

Would be interested in consoling another comparison test on the geeky side. One in which two comparably equiped systems (or three) are rigged to an Inverter,or run from a Backup Power supply. Running the same apps. Noting which system ran out of "juice" sooner.

You could possibly say that if for example the 90 watt/130 watt killed the battery/backup sooner,although it did the applications over x amount of times (greater).

But what I am looking at is simply if at all possible to run the test. Perhaps with a timer hooked onto the battery connection to measure the juice left,and/or time battery konked out.

The H.264 test would seem to be the better test. Since it would be likely that using such a connection scenario might be most utilized.

Notice that most of the i7 core systems though,run from extremely large power supply systems. Though I have seen them run from as low as 650 watt from articles. Also note that the formulas used in your test should be calculable via math. However you might notice that there is a formula which decides the length of the power supplies capability.

Of course you would not be able to watch a movie in x amount of greater frame rate. Or in fast forward mode. Then the Inverter I have (which I have used to start my auto at times) might not even run a system that needed a cold jump at x amount of watts. X amount of Amps. I guess joules is a derivative of Amps. I could look it up,but I'll just say I dont know here.
Reply

if one already owns a C2Q, undervolting it is a nice option. My Q9450 runs perfectly stable @ 3.4GHz wo 1.184V, I don't care much about the energy reduction but the heat reduction. I can have a already silent fan running at only 600rpm on my True, and that means a nice and silent workplace :)

Normally, with higher volts it is stable at 3.7GHz(and hot) but that ain't worth it because the sound from the fans for just a little more performance.

My E8400 undervolts even better, it runs a Linux home-server 24/7 @ only 0.91V and w both bridges undervolted. Reply

It's closer to $150 more than an equivalent C2Q setup. X58 boards are now available for about $200 ($100 more than UD3R for example) and there's a 6GB DDR3-1333 kit for $110 shipped (about $70 higher than a 4GB DDR2 kit - remember, you're getting more memory). Reply

It seems like you should have included an e8400 @ 3GHz as a direct comparison to the Q9650 and PhII 940. Yes, the work would have taken longer - but the chip also runs a lot lower on power, I would imagine. And it would have been interesting to me to see how efficient those chips are compared to the quads available today. Reply

The big problem with this review is that the power consumption setup is based on an inappropriate premise: That someone will use these chips in a high-end rig.

These chips are *NOT* for someone who is running a GTX 280. I mean, that video card draws more power at idle than these chips draw at full load.

No, these chips are for thin-and-light desktops that otherwise have too small a thermal envelope for quad-core. In *THOSE* systems, the extra money is worth it, since the comparison is these new chips, or dual-core. Not these chips vs. the 95W quad-cores.

For example, Intel has two mini ITX motherboards (DG45FC and DQ45EK) that have a max TDP of 65W. Those are the target for these new chips. For ultra-small systems without a 200+W video card. Reply

The Q9550S and the other 65W quad-cores are designed for OEMs or anyone trying to cram as much power into a very small space. I’d expect that these CPUs would be better suited for something like an iMac rather than a normal sized desktop. The problem is that in a normal desktop you’ve got more than enough room to keep even a Core i7 cool, but in some of these OEM designs (like the iMac or Dell XPS One 24) there’s hardly enough room for a normal heatsink and fan. Reply

some results look strange like x264
Phenom2, q9400 and q9450 have the same speed and same average power.
Shouldn't they have the same total energy comsumption?
I would say there's something wrong in this chart.

BeHardware reviewed the q9300 awhile back and tested consumption directly from the voltage rail and found it never pulled more than about 45 watts full load. This is consistent with the fact that the q9300 and its q8000 series cousins typically pull about 30W+ less than the q6600. Intels own processor information finder shows identical voltages for the 9300 and this "S" series CPU and the voltages are identical.. Only thing different is the stepping. Could the q9300 have been labeled as a 50W part had Intel cared to do so? Or even a 65W for that matter (giving margin of yield variation)?

You're right that the Core i7 is overall a more efficient machine. However, the S series are LGA775 chips, which means that could fall into the upgrade path for many users, or simply be the chip of choice for someone interested in an $85 board and power friendly chip. It's the added premium of X58 and triple channel DDR3 memory that sets th Core i7 cost outside of the mainstream, not just the cost of the chip itself. Otherwise I'd be looking at a Core i7 machine right now.... Reply

I have a hard time wrapping my mind around a low(er) TDP part that cannot OC better than a non low TDP-binned processor? Is it just that the one you have for testing is a uniquely "gifted" cpu? It's my understand (correct me if I'm wrong) that lower TDP parts are due to superior construction that reduces leaks that would otherwise just go to heat production. If this is the case I can't see why a lower TDP part couldn't OC higher....

Please enlighten me as it doesn't make a lot of sense. Oh and thanks for the review because I'm sure I'm not the only one that just assumed a lower TDP part would on average OC better than a higher TDP part. Reply

Lower power parts can come from a variety of things, but binning is a major impact. In many cases, chips that need less voltage run cooler and OC better; likewise, chips that *can* run with less voltage can also run fine at higher speeds with a "normal" voltage. That puts Intel (and AMD) in the position of binning for several options: do you want a low power chip, a high default MHz chip, or some combination of the two?

My guess would be that a lot of the QX9650/QX9770 and similar "extreme" CPUs can run at lower voltages and lower MHz. However, Intel might take CPUs that bin successfully at high clocks first, and then bin the remaining to see if some can run at lower voltage, which could then account for S-series chips that don't OC any better than regular chips.

Also remember that overclocking depends on many areas of the CPU, so it could be that these low voltage chips also tend to have limitations elsewhere -- maximum OC will always depend on the weakest link in the chip, which is why we see variability. Perhaps this being an ES chip has an impact, or it could just be random luck. Good or bad luck, that's the question.

I'm sure some S-series chips will overclock slightly better than regular Penryns, but at the end of the day the last 100-400 MHz really only matters to people running competitive benchmarks. Heck, I'm still running a Q6600@3.30GHz, and I have never encountered any situation where I think, "I wish my CPU had a little more oomph...." YMMV, naturally. Reply

Anand, can you provide any further explanation why Intel’s power gate transistors gave the i7 a higher energy efficiency on the Fallout 3 test in terms of Joules but didn't actually give it a lower "at the wall" power draw in average watts? I'm just not following why it's making any difference in joules when it's not in watts (compared to the P2 940) if the dt is the same.

Also, any interest in adding testing of the 50 watt Xeon L5420 to the benchmark database? Intel has had low voltage parts for almost a year, they just haven't brought them into the consumer products until now. For the most part they've been ignored by review sites. Reply

About the temperature advantage, I think the results are not trustworthy enough because there is no software that can accurately measure the absolute temperature of the cores.
Even though Intel disclosed the tjunction max spec for their 45nm parts, they also said that the measure is not completely precise because each chip has a different tj max temperature that can not be read from any register.

Only i7's tj max can be obtained directly from the chip, so that the absolute core temperature can be better calculated.

Also on Intel's presentation, they stated that the DTS' readings are not accurate enough for measuring idle temps. Anyway this doesn't apply to this article as you're comparing mainly load temps, not idle ones, and when under load and the more the on die temp approaches tj max, the more accurate the DTS' reading becomes.

And as the Tcase temp is read and by a motherboard's sensor (at least AFAIK from the available Intel's docs), it isn't completely reliable either.

Aside from that, excellent article. Not many sites measure the "performance per watt" factor. The only thing I'd add to this article, is a direct comparison with a non-s 9550 e0 chip, specially in the overclocking department, as I tend to think that "cooler chips" will be actually most wanted by overclockers than by energy savers (mainly considering the high price premium)

Those processors are perfect as replacements in servers already validated for the 95W version of the same processor. While buying i7 would be better, maybe the i7 servers weren't validated (remember the 3 years of support for business-related hardware lines) Reply

How about buying a regular part, and under volting it yourself ? Can these new parts be under volted themselves ? Why isnt intel selling a Core 2 that is internally a laptop CPU, but externally a socket 775 part ? e.g. "we" would like a desktop 25W TDP CPU.

Things that make you go "hmmm" ?

Too bad Cyrix, and possibly more ( Centuar ? ) CPU makers are no longer around, we're getting stuck in some stupid war between two manufactures, instead of embracing newer, and better technology; At Intel's whim. How about something truly revolutionary . . . Reply

A few kW average per month is most people's electric bills. :) I wanted to say, if more people considered every watt used a big deal then more people might be able to run off panels on their roof. Every watt certainly matters by the time you count how many electrical devices you have in your home. Reply

Intel's pricing is still so far off base on their quad cores. Right now on the egg a Phenom II X4 920 @ 2.8GHz is $195. The closest Intel CPU, the Q9550 Yorkfield @ 2.83GHz is $281.99. Granted, the Intel CPU is ever so slightly faster, and may overclock a bit better, but almost $100 worth of better? At the higher end the price gap becomes utterly ridiculous. The Phenom II X4 940 Black Edition @ 3.0GHz is $235. The closest Intel CPU, that also has an unlocked multiplier like the AMD, is the Core 2 Extreme QX9650 @ 3.0GHz which costs 1,029.99. Again, granted, the Intel CPU may be slightly faster and a better overclocker, but that is a price difference of 794.99. I could build an entire midrange system for that. What I would much rather see as a consumer, rather that a processor that consumes a bit less power, is a quad that is actually reasonably priced. While this seems to be the direction AMD has gone, Intel still charges you and arm a leg just for entry to the quad club. I'm no AMD fanboy, but I do believe in buying what gives the biggest bang for the buck. I bought an lga 775 motherboard almost a year ago because AMD simply couldn't compete at the time. I also had the intention of going to a quad core when they became more mainstream and affordable. If the prices of Intel chips don't drop substantially soon, I may just go ahead and get an AMD product and use the money I save to get a new motherboard. Reply

Are you even bothering to look at the benchmarks? I understand that it is easiest to compare a PIIX4 940 Black @ 3.0GHz with a C2E QX9650 @ 3.0GHz because the numbers are the same. But looking at the x264 benchmarks(the only one that is relevant to me) it's really only as fast as the Q9400 @ 2.66GHz. According to the egg, the X4 940 is on sale for $253 while the Q9400 is only $229.

Your entire argument is pointless because you've fallen for the MHz myth it seems. There hasn't been clock parity since the days of PIII and original Athlon. Reply

Yes, I am considering benchmarks. According to Anand techs own video game benchmarks (the only thing I care about) in the recent Phenom II article, the closest match to the Phenom II X4 920 in terms of performance is the Q9450, which is no longer listed on the egg. However the Q9550, which is still listed, is anywhere from about 2 to 10 fps faster in benchmarks, but costs almost $100 more. I am not trying to argue that these two CPUs have identical performance. As I stated previously the Intel CPU is without question a faster CPU, but it is not all that much faster especially when you consider the price difference. That boils down to anywhere from about $9 to $50 per additional fps, depending on the game. I'm sorry but the cost to benefit simply is not there. You would be much better off investing that money into a better video card, motherboard, or more RAM. This isn't even factoring in that AMD motherboards tend to be cheaper as well, lowering the overall cost of the system, or conversely, allowing you to get a better AMD CPU to make up that performance gap and still end up ahead. Reply

Far and away the most cost efficient thing for those like you who only care about games is to stick with your dual core, followed by getting a faster dual-core or an SSD. Check the performance numbers in the Phenom II launch article, as most of the games are not using the quad-cores, very few of them look good in added performance for the money. Reply

What it really boils down to is "which socket do you already have?". And Intel knows this. Think about it, they've been beating AMD pretty much since the C2D's came out so most people adopted that platform and already own Socket T/775 boards with C2D chips. Now that AMD has released a competing 45nm quad core, Intel has had to lower their prices to attract purchases of customers they already had. The decision for many people (including yourself) is do you spend $280 on a new Q9550 or $195 for a new P2 920 and another $100-150 on a new motherboard that will make it work? Clearly the almost $300 CPU is the cheaper option. And I'd bet most will go for the Q8200 or the Q9400. "Most" being the average consumer that buys it and gets someone else to install it for them. Intel would be silly not to get another $85, or even the $35 for the Q9400, out of customers that are upgrading from Core2Duo's. IMO, it's worth it just to avoid the hassle of changing 2 pieces of hardware at 1 time. That's a major headache if one of the parts is DOA.

The only place Intel can lose business (in reference to quad cores) to AMD right now is people buying Phenom 9600s for $110 that want a quad core "just because", to people that already have an AM2 platform, AMD fanbois, and people that have not bothered to upgrade their 2Ghz P4 yet. And those last aren't going to go with AMD because they think AMD parts are cheap for a reason. There are many people that just won't buy from tech companies like AMD because they consider them "second rate" businesses with low quality products, even though they've never owned anything they make. Reply

You people really need to read before you post. At the bottom of page one the author states and I quote "...and in case you’re wondering, no, they don’t overclock any better. Our Q9550S couldn’t get any further than the Q9550 we used in our Phenom II review." If you aren't going to even bother reading the article why do you bother to post and ask questions that were answered in the article you clearly didn't even read? Reply

In the Photoshop tests I'm assuming you calculated joules by multiplying the time it took to complete the test by the average system wattage drawn from the wall. If that is the case the Phenom II 940 joules should be closer to about 3800 joules (24.2sec * 157watts) and not the 4700 joules you have listed.

In general when people say average, they are talking about the Arithmetic mean.
Arithmetic mean = 1/n*(X1 + ... + Xn)
or in English a list of numbers divided by the number of items in the list.

In your case this would mean summing the list of power measurements taken at one second intervals and dividing by the number of measurements which would be the integer number of seconds. You could then calculate joules by multiplying that average by the total time.

The only way the sum of power measurements and calculation made by multiplying the average power by the time would be different is if the number of measurements for the sum and the average are different. In this scenario, the calculation made with more data points would be more accurate (think integration).

So the question becomes: How did you calculate your average? It appears that your average has more data points given that the total test time is measured to 1/10 seconds and your summation was only one second intervals.

That said, the difference between the summation and the result calculated from the average should be small as you stated. The Q9550S results, for instance, only differs by 113(3244-3131) joules and the Core i7-920 differs by a mere 86(2818-2732) joules. Even the Phenom X4 9950 only has a delta of 69(5474-5405) joules. However, the Phenom II 940 has a delta of 898(4697-3799) joules.

This massive difference leads me to suspect that either the average power, the total time, or the total energy for this processor was reported incorrectly. If we assume the average power and the maximum power are the same, then the delta shrinks to 220(4697-4490) joules. Alternately, if we assume that the Phenom II 940 is the same speed as the Phenom 9950, the delta shrinks to 207(4697-4477) joules. Both of the assumptions seem unreasonable to me, and neither get the delta as small as it should be. So I ask, now that I've presented a reasonable case, please recheck your total energy numbers as Ryun suggested. Reply

Woops, you're completely right :) The issue wasn't with the power measurement but with the performance. The performance data for the run that I measured power under was incorrect. A re-run fixes that problem. The Q9450 was also impacted slightly.

It's worth mentioning that the performance and power data are taken at two different times. First the performance data, then the power data. The performance during the power run is close but not always identical to the performance during the performance run. There's going to be some variation depending on the test.

I have to agree. There is something wrong with Anand's methodology. Also, look at his specious reasoning for the difference in processor ranking between his "average" power and the energy consumed in the Fallout 3 section, where the tests are run for the same time interval. He is measuring total system power, so the improved idle efficiency of the Nehalems should already be incorporated in those numbers. Average power draw is by definition total energy consumed divided by time interval over which it is consumed. Either taking instantaneous measurements and treating them as averages for each second or simple human error could be responsible for the discrepancy. Reply

I wouldn't call it suspicious, just a flaw in the procedure. If you have a sine wave and a cosine wave at the same frequency, amplitude, and offset measured once per period, one will look much larger than the other even though they average out to be exactly the same. Likewise, if you have two computers drawing the same average power, but you happen to record one during mostly high fluctuations and the other during mostly low fluctuations, you'll get two very different results.
You need more samples to get accurate results. The best method would be to record a power graph using the smallest period possible. Then, integrate the power under the curve. Convert the units to seconds to get energy in joules. Divide by the number of samples to get the average power. Reply

JPForums, I appreciate your efforts to elucidate my remarks to Anand, but I must comment on two things. First, the word I used was "specious" not "suspicious". There is a difference, just as there is a difference between "average power draw" and "an average of periodically sampled power draws". These two are only guaranteed to coincide if the samples themselves are average powers or in the limit as the period they are sampled over approaches 0. (The latter remark is directed at your definition of average in the first paragraph of your other post). Reply